Chain Rule Lemma

Question

This should be really simple, but I am not following where the Chain Rule is placed here.

If we set $f(tx_1, \dots, tx_n) = f(h_i(x_1,t), \dots, h_n(x_n,t))$, the total derivative is $$df = \sum \frac{\partial f}{\partial h_i} \frac{\partial h_i}{\partial t} = \sum x_i\frac{\partial f}{\partial h_i} $$. In the last step, as we setting $x_i = 1$?

If $f$ is smooth, wouldn't that imply all derivatives are smooth ($C^\infty$ smooth) and defined in the first placed and hence $g_i$ must exist no matter what point we are at?

Answer 1

The second equality in the second displayed equation is true because of the chain rule. The functions $g_i$ so defined in the last line of the proof then satisfy the conclusion of the theorem (due to the second displayed equation).

It's true that $f$ is smooth and it's derivatives exist, but you want $g_i$ that specifically satisfy the first displayed equation. The initial condition on $f$ is for the fundamental theorem of calculus, used in the proof without comment.

Answer 2

Recall the chain rule.

$$\frac{\partial f(x(t))}{\partial t} = \sum_{i=1}^n \frac{\partial f(x_i(t))}{\partial x_i} \frac{\partial x_i(t)}{\partial t} $$

Here $x_i(t) = tx_i $ and $ x(t) = (tx_1, ... tx_n)$

And so $$\frac{\partial x_i(t)}{\partial t}= x_i$$